Why does merge result in more rows than original data?

Question

When I merge two data frames, the result has more rows than the original data.

In this instance, the all dataframe has 104956

Answer 1

First, from ?merge:

The rows in the two data frames that match on the specified columns are extracted, and joined together. If there is more than one match, all possible matches contribute one row each.

Using your link in the comments:

url    <- "http://koeppen-geiger.vu-wien.ac.at/data/KoeppenGeiger.UScounty.txt"
koppen <- read.table(url, header=T, sep="\t")
nrow(koppen)
# [1] 3594
length(unique(koppen$FIPS))
# [1] 2789

So clearly koppen has duplicated FIPS codes. Examining the dataset and the website, it appears that many of the counties are in more than one climate class, so for example, the county of Ankorage, Alaska has three climate classes:

koppen[koppen$FIPS==2020,]
#     STATE    COUNTY FIPS CLS  PROP
# 73 Alaska Anchorage 2020 Dsc 0.010
# 74 Alaska Anchorage 2020 Dfc 0.961
# 75 Alaska Anchorage 2020  ET 0.029

The solution depends on what you are trying to accomplish. If you want to extract all rows in all with any FIPS that appear in koppen, either of these should work:

merge(all,unique(koppen$FIPS))

all[all$FIPS %in% unique(koppen$FIPS),]

If you need to append the county and state name to all, use this:

merge(all,unique(koppen[c("STATE","COUNTY","FIPS")]),by="FIPS")

EDIT Based on the exchange below in the comments.

So, since there are sometimes multiple rows in koppen with the same FIPS, but different CLS, we need a way to decide which of the rows (e.g., which CLS) to pick. Here are two ways:

# this extracts the row with the largest value of PROP, for that FIPS
url        <- "http://koeppen-geiger.vu-wien.ac.at/data/KoeppenGeiger.UScounty.txt"
koppen     <- read.csv(url, header=T, sep="\t")
koppen     <- with(koppen,koppen[order(FIPS,-PROP),])
sub.koppen <- aggregate(koppen,by=list(koppen$FIPS),head,n=1)
result     <- merge(all, sub.koppen, by="FIPS")

# this extracts a row at random
sub.koppen <- aggregate(koppen,by=list(koppen$FIPS), 
                        function(x)x[sample(1:length(x),1)])
result     <- merge(all, sub.koppen, by="FIPS")